Rank-nullity theorem — proof — Hinglish notes

4.5.14 · Maths › Linear Algebra (Full)

WHAT is the theorem?

$ker T$ , $V$ ka subspace hai (the input side).
$im T$ , $W$ ka subspace hai (the output side).
Dhyan raho dono numbers milke $dim V$ dete hain — domain ki dimension, codomain $W$ ki nahi.

WHY should it be true? (rigour se pehle intuition)

HOW to prove it — derivation from scratch

Hum ise carefully build karte hain. Key trick hai Basis Extension Theorem: kisi bhi subspace ke basis ko poore space ke basis tak extend kiya ja sakta hai.

Formula Proof

Setup. Maano $dim V = n$ . Maano $nullity (T) = k$ , toh kernel ka ek basis chuno: ${u_{1}, \dots, u_{k}} a basis of ker T .$

Step 1 — Extend. Kyunki $ker T \subseteq V$ hai, ise poore $V$ ke basis tak extend karo: ${u_{1}, \dots, u_{k}, v_{1}, \dots, v_{n - k}} a basis of V .$ Yeh step kyun? Basis-extension theorem guarantee karta hai ki hum $n - k$ vectors add karke $V$ ka full basis bana sakte hain. Yeh "kernel part" ko "baaki sab" se saaf taur par alag kar deta hai.

Claim: $B = {T (v_{1}), \dots, T (v_{n - k})}$ is a basis of $im T$ . Agar hum yeh prove kar lein, tab $dim (im T) = n - k$ hoga, aur $rank + nullity = (n - k) + k = n = dim V . ■$

Toh hume dikhana hai ki $B$ (a) spans $im T$ aur (b) is linearly independent.

Step 2 — $B$ spans $im T$ . Koi bhi $w \in im T$ lo, toh $w = T (v)$ kisi $v \in V$ ke liye. $v$ ko full basis mein likho: $v = \sum_{i = 1}^{k} a_{i} u_{i} + \sum_{j = 1}^{n - k} b_{j} v_{j} .$ $T$ apply karo (linear hai), aur $T (u_{i}) = 0$ use karo kyunki har $u_{i} \in ker T$ : $w = T (v) = \sum a_{i} = 0 T (u_{i}) + \sum b_{j} T (v_{j}) = \sum_{j = 1}^{n - k} b_{j} T (v_{j}) .$ Yeh step kyun? Har output sirf survivors $T (v_{j})$ se banta hai — kernel directions kuch contribute nahi karti. Isliye $B$ , image ko span karta hai. ✓

Step 3 — $B$ is linearly independent. Maano $\sum_{j = 1}^{n - k} c_{j} T (v_{j}) = 0$ . Linearity se: $T (\sum_{j} c_{j} v_{j}) = 0 ⟹ \sum_{j} c_{j} v_{j} \in ker T .$ Yeh step kyun? Agar survivor-outputs ka koi combination zero hai, toh inputs ka matching combination kernel mein hai — humhe wapas $u_{i}$ par le jaata hai.

Kyunki ${u_{1}, \dots, u_{k}}$ is a basis of $ker T$ , hum likh sakte hain $\sum_{j} c_{j} v_{j} = \sum_{i} d_{i} u_{i} ⟹ \sum_{j} c_{j} v_{j} - \sum_{i} d_{i} u_{i} = 0.$ Lekin ${u_{1}, \dots, u_{k}, v_{1}, \dots, v_{n - k}}$ is a basis of $V$ , isliye linearly independent hai, toh saare coefficients zero ho jaate hain: $c_{1} = \dots = c_{n - k} = 0 (aur d_{i} = 0) .$ Yeh step kyun? $V$ ke poore basis ki independence $c_{j}$ ko zero hone par majboor karti hai — yahi hume chahiye tha. ✓

Dono (a) aur (b) hold karte hain, toh $B$ , $im T$ ka ek basis hai jiska size $n - k$ hai. Ho gaya.

Worked examples

Common mistakes (steel-manned)

Recall Feynman: 12-saal ke bachche ko samjhao

Ek machine socho jisme 3 input levers hain (3 dimensions of input). Jab tum levers push karte ho, kuch pushes kuch nahi karte — machine unhe ignore kar deti hai (woh "kernel" levers hain). Jo pushes kuch karte hain woh output mein movement karte hain. Badi baat yeh hai: har lever ya toh ignore hoti hai ya ek unique naya output banati hai. Toh (ignored levers) + (useful levers) = (total levers). Bas yahi poora theorem hai!

Active-recall flashcards

#flashcards/maths

Rank–nullity theorem kya kehta hai?

Ek linear map

T : V \to W

ke liye jahan

V

finite-dimensional hai,

rank (T) + nullity (T) = dim V

Rank kis space ki dimension hai?

T

ki image (range) ki, yaani

dim (im T)

Nullity kis space ki dimension hai?

T

ke kernel (null space) ki, yaani

dim (ker T)

Sum domain ki dimension ke barabar hota hai ya codomain ki?

Domain

V

ki.

Kaun sa theorem kernel basis ko

V

ke basis tak extend karne deta hai?

The basis-extension theorem.

Proof mein kaun sa set image ka basis dikhaya gaya hai?

{T (v_{1}), \dots, T (v_{n - k})}

, yaani kernel ke bahar ke basis vectors ki images.

T (v_{j})

image ko kyun span karte hain?

Koi bhi

v

, kernel part (jo 0 par map hoti hai) aur

v_{j}

part mein split hoti hai, toh har output

T (v_{j})

ka combination hai.

T (v_{j})

independent kyun hain?

Agar

\sum c_{j} T (v_{j}) = 0

toh

\sum c_{j} v_{j} \in ker T

; use

u_{i}

se likhne par aur full-basis independence use karne par saare

c_{j} = 0

ho jaate hain.

Rank 1 wale

2 \times 3

matrix ki nullity kya hai?

3 - 1 = 2

Kya vectors ki independence linear map se hamesha preserve hoti hai?

Nahi; ek map independent vectors ko collapse kar sakti hai — isliye proof kernel ke zariye argue karta hai.

Connections

Kernel and Image of a Linear Map
Basis Extension Theorem
Dimension of a Vector Space
Rank of a Matrix (row vs column space)
Invertible Linear Maps and Isomorphisms (square map injective $⟺$ surjective via rank–nullity)
Solving Linear Systems — free variables count

Rank-nullity theorem — proof